Head and summary of the Nhanes 2015/2016 dataset

library(knitr)
library(kableExtra)

#First 6 rows of the data set prior to data wrangling.
knitr::kable(head(nhanes.2015.2016, format="latex", booktabs=TRUE)) %>% 
  kable_styling(latex_options="scale_down")
SEQN ALQ101 ALQ110 ALQ130 SMQ020 RIAGENDR RIDAGEYR RIDRETH1 DMDCITZN DMDEDUC2 DMDMARTL DMDHHSIZ INDFMPIR BPXSY1 BPXDI1 BPXSY2 BPXDI2 BMXWT BMXHT BMXBMI BMXLEG BMXARML BMXARMC BMXWAIST HIQ210
83732 1 NA 1 1 1 62 3 1 5 1 2 4.39 128 70 124 64 94.8 184.5 27.8 43.3 43.6 35.9 101.1 2
83733 1 NA 6 1 1 53 3 2 3 3 1 1.32 146 88 140 88 90.4 171.4 30.8 38.0 40.0 33.2 107.9 NA
83734 1 NA NA 1 1 78 3 1 3 1 2 1.51 138 46 132 44 83.4 170.1 28.8 35.6 37.0 31.0 116.5 2
83735 2 1 1 2 2 56 3 1 5 6 1 5.00 132 72 134 68 109.8 160.9 42.4 38.5 37.7 38.3 110.1 2
83736 2 1 1 2 2 42 4 1 4 3 5 1.23 100 70 114 54 55.2 164.9 20.3 37.4 36.0 27.2 80.4 2
83737 2 2 NA 2 2 72 1 2 2 4 5 2.82 116 58 122 58 64.4 150.0 28.6 34.4 33.5 31.4 92.9 NA 
#Summary of data set after data wrangling
summary(nhanes2)
##       SEQN           Alcohol_Year       Smoked_100      Gender    
##  Min.   :83732   Don't know:   3   Don't know:   8   Female:2976  
##  1st Qu.:86164   No        :1728   No        :3406   Male  :2759  
##  Median :88668   Yes       :3477   Refused   :   2                
##  Mean   :88679   NA's      : 527   Yes       :2319                
##  3rd Qu.:91178                                                    
##  Max.   :93702                                                    
##                                                                   
##       Age                      Race        Education    
##  Min.   :18.00   Black           :1227   Min.   :1.000  
##  1st Qu.:32.00   Mexican American:1018   1st Qu.:3.000  
##  Median :48.00   Other Hispanic  : 750   Median :4.000  
##  Mean   :48.05   Other Race      : 901   Mean   :3.442  
##  3rd Qu.:63.00   White           :1839   3rd Qu.:4.750  
##  Max.   :80.00                           Max.   :9.000  
##                                          NA's   :261    
##              Marital_Status Household_Size Income_to_Pov    Systolic_BP1  
##  Married            :2780   1: 770         Min.   :0.000   Min.   : 82.0  
##  Never Married      :1004   2:1546         1st Qu.:1.060   1st Qu.:112.0  
##  Divorced           : 579   3:1037         Median :1.980   Median :122.0  
##  Living with partner: 527   4: 936         Mean   :2.403   Mean   :125.1  
##  Widowed            : 396   5: 699         3rd Qu.:3.740   3rd Qu.:134.0  
##  (Other)            : 188   6: 379         Max.   :5.000   Max.   :236.0  
##  NA's               : 261   7: 368         NA's   :601     NA's   :334    
##  Diastolic_BP1     Systolic_BP2   Diastolic_BP2        BMXWT       
##  Min.   :  0.00   Min.   : 84.0   Min.   :  0.00   Min.   : 32.40  
##  1st Qu.: 62.00   1st Qu.:112.0   1st Qu.: 62.00   1st Qu.: 65.90  
##  Median : 70.00   Median :122.0   Median : 70.00   Median : 78.20  
##  Mean   : 69.52   Mean   :124.8   Mean   : 69.35   Mean   : 81.34  
##  3rd Qu.: 78.00   3rd Qu.:134.0   3rd Qu.: 78.00   3rd Qu.: 92.70  
##  Max.   :120.00   Max.   :238.0   Max.   :144.00   Max.   :198.90  
##  NA's   :334      NA's   :200     NA's   :200      NA's   :69      
##      BMXHT           BMXBMI          BMXLEG         BMXARML     
##  Min.   :129.7   Min.   :14.50   Min.   :26.00   Min.   :28.20  
##  1st Qu.:158.7   1st Qu.:24.30   1st Qu.:36.00   1st Qu.:35.20  
##  Median :166.0   Median :28.30   Median :38.60   Median :37.10  
##  Mean   :166.1   Mean   :29.38   Mean   :38.58   Mean   :37.15  
##  3rd Qu.:173.5   3rd Qu.:33.00   3rd Qu.:41.20   3rd Qu.:39.00  
##  Max.   :202.7   Max.   :67.30   Max.   :51.50   Max.   :47.40  
##  NA's   :62      NA's   :73      NA's   :390     NA's   :308    
##     BMXARMC         BMXWAIST          HIQ210     
##  Min.   :17.10   Min.   : 58.70   Min.   :1.000  
##  1st Qu.:29.50   1st Qu.: 87.60   1st Qu.:2.000  
##  Median :32.70   Median : 98.30   Median :2.000  
##  Mean   :33.11   Mean   : 99.57   Mean   :1.915  
##  3rd Qu.:36.20   3rd Qu.:109.30   3rd Qu.:2.000  
##  Max.   :58.40   Max.   :171.60   Max.   :9.000  
##  NA's   :308     NA's   :367      NA's   :1003   
##                          EducationX     agegroup    Income_Group
##  < 9th grade                  : 655   18-29 :1192   1   :1179   
##  9-11th grade                 : 643   30-39 : 933   2   :1406   
##  High school graduate         :1186   40-49 : 913   3   : 876   
##  Some college/Uni             :1621   50-59 : 888   4   : 486   
##  College/Uni graduate or above:1366   60-69 : 917   5   : 362   
##  9                            :   3   70-80+: 892   NA's:1426   
##  NA's                         : 261

3 functions created (saved in lib file).

  1. ‘corr_func’ to quickly test correlations on the variables whilst exploring the data). The sole purpose using this rather than the built-in function is so the argument ‘complete.obs’ doesn’t have to be typed each time.
  2. ‘grp_func’ to quickly allow the user to input a variable to group by and a variable to analyse which will output the min, max, and mean.
  3. ‘prop_func’ takes the following aruguments: dataframe, catagorical group variable, and binary variable to analyse. Then outputs a new dataframe of proportions based on these arguments (split by gender). This will be used in the shiny dashboard.

Below code to firstly find out all the columns containing numeric data (numeric_vars. Then a new data table is created with these values (numeric_data). From this table, all the correlations were worked out wit the ‘cor’ function and saved under a variable names ‘correlations’. A correlation plot was created to clearly show any positive or negative correlations between these variables.

Group by agegrp to look at proportions of people who have and have not smoked 100 cigarettes in their. Assume that the the older generations will have a larger proportion of smokers than the youngers ones. Displayed in pie charts

smoked = prop_func(nhanes2, 'agegroup', 'Smoked_100', percentage = FALSE)

p1 = smoked[1:2,]
p2 = smoked[3:4,]
p3 = smoked[5:6,]
p4 = smoked[7:8,]
p5 = smoked[9:10,]
p6 = smoked[11:12,]

pie1 = ggplot(p1, aes(x='', y=Total_Proportion, fill =Smoked_100 )) + 
  geom_bar(stat='identity', width=1) + labs(title = 'Age - 18-29') + 
  coord_polar('y', start=0) + 
  geom_text(aes(label = percent(Total_Proportion)), position = position_stack(vjust = 0.5), size = 4) +
  scale_fill_brewer(palette = 'Dark2') +
  theme_void()

pie2 = ggplot(p2, aes(x='', y=Total_Proportion, fill =Smoked_100 )) + 
  geom_bar(stat='identity', width=1) + labs(title = 'Age - 30-39') + 
  coord_polar('y', start=0) + 
  geom_text(aes(label = percent(Total_Proportion)), position = position_stack(vjust = 0.5), size = 4) +
  scale_fill_brewer(palette = 'Dark2') +
  theme_void()

pie3 = ggplot(p3, aes(x='', y=Total_Proportion, fill =Smoked_100)) + 
  geom_bar(stat='identity', width=1) + labs(title = 'Age - 40-49') + 
  coord_polar('y', start=0) + 
  geom_text(aes(label = percent(Total_Proportion)), position = position_stack(vjust = 0.5), size = 4) +
  scale_fill_brewer(palette = 'Dark2') +
  theme_void()

pie4 = ggplot(p4, aes(x='', y=Total_Proportion, fill =Smoked_100)) + 
  geom_bar(stat='identity', width=1) + labs(title = 'Age - 50-59') + 
  coord_polar('y', start=0) + 
  geom_text(aes(label = percent(Total_Proportion)), position = position_stack(vjust = 0.5), size = 4) +
  scale_fill_brewer(palette = 'Dark2') +
  theme_void()

pie5 = ggplot(p5, aes(x='', y=Total_Proportion, fill =Smoked_100)) + 
  geom_bar(stat='identity', width=1) + labs(title = 'Age - 60-69') + 
  coord_polar('y', start=0) + 
  geom_text(aes(label = percent(Total_Proportion)), position = position_stack(vjust = 0.5), size = 4) +
  scale_fill_brewer(palette = 'Dark2') +
  theme_void()

pie6 = ggplot(p6, aes(x='', y=Total_Proportion, fill =Smoked_100)) + 
  geom_bar(stat='identity', width=1) + labs(title = 'Age - 70-80+') + 
  coord_polar('y', start=0) + 
  geom_text(aes(label = percent(Total_Proportion)), position = position_stack(vjust = 0.5), size = 4) +
  scale_fill_brewer(palette = 'Dark2') +
  theme_void()


grid.arrange(
  pie1, pie2, pie3, pie4, pie5, pie6, nrow=2, ncol = 3, top = 'Proportion of people who have smoked 100 cigarettes'
)

Next test to see if someone with a higher education level is less likely to have smoked 100 cigarettes in their life. From the results we can see that as a total proportion between male and female there is not much difference. However if we look at only the males, we see that a college/uni graduate is less likely to have smoked 100 than someone with a lower education.

ed_smoke = prop_func(nhanes2, "EducationX", "Smoked_100")
knitr::kable(ed_smoke)
EducationX Smoked_100 Male_Proportion Female_Proportion Total_Proportion
< 9th grade No 46.71% 79.1% 64.0%
< 9th grade Yes 53.29% 20.9% 36.0%
9-11th grade No 33.0% 61.2% 45.79%
9-11th grade Yes 67.0% 38.8% 54.21%
High school graduate No 38.6% 62.1% 50.169%
High school graduate Yes 61.4% 37.9% 49.831%
Some college/Uni No 44.4% 62.7% 54.66%
Some college/Uni Yes 55.6% 37.3% 45.34%
College/Uni graduate or above No 64.4% 78.8% 71.9%
College/Uni graduate or above Yes 35.6% 21.2% 28.1%

Below some box plots and scatter plots have been made influenced by the correlation plot. We hope to find some interetsing results from this. cor_func and grp_func was also used for futher testing. Based on the results some furhter hypothesis testing will be carried out.

b1 = ggplot(nhanes2, aes(x=agegroup, y=Systolic_BP1, color=Gender)) + 
  geom_boxplot()
b1

b1_test = filter(nhanes2, Age>69)
b1_func = grp_func(b1_test, 'Gender', 'Systolic_BP1')
knitr::kable(b1_func)
Gender Mean Max Min
Female 139.9763 210 84
Male 136.5012 204 92

Based on the above information 2 hypothesis tests will be carried out. 1 to see if men of all ages have higher Systolic BP than females, and another to test if females aged 70+ have a higher Systolic BP than men. (teststing at bottom of repoort)

#Filter dataframe to remove below value
df2 = filter(nhanes2, EducationX !='9')
b2 = ggplot(df2, aes(x=EducationX, y=Income_to_Pov, color=Gender)) + 
  geom_boxplot()
b2

nhanes2$Household_Size = as.factor(nhanes2$Household_Size)
b3 = ggplot(nhanes2, aes(x=Household_Size, y=Income_to_Pov, color=Gender)) + 
  geom_boxplot()
b3

Based on the above plot, a hypothesis test will be carried out to see if females living alone earn more than men living on their own. (test below)

b4 = ggplot(nhanes2, aes(x=Race, y=Income_to_Pov, color=Gender)) + 
  geom_boxplot()
b4

#Scatter plots
#Filter dataframe to remove below values
df1 = filter(nhanes2, Smoked_100 !="Don't know" & Smoked_100 !='Refused')

s1 = ggplot(df1, aes(x=Systolic_BP1, y=Systolic_BP2, color = Smoked_100)) + 
  geom_point(alpha=0.2) + 
  scale_color_manual(breaks = c('Yes', 'No'), values = c('blue', 'red'))
s1

corr_func('Systolic_BP1', 'Systolic_BP2')
## [1] 0.9622873
s2 = ggplot(nhanes2, aes(x=BMXWT, y=Systolic_BP1, color=Gender)) + 
  geom_point(alpha=0.2) +
  scale_color_manual(breaks = c('Male', 'Female'), values = c('blue', 'red'))
s2

corr_func('BMXWT', 'Systolic_BP1')
## [1] 0.1225117
grp_func(nhanes2, 'Gender', 'BMXBMI')
## # A tibble: 2 x 4
##   Gender  Mean   Max   Min
##   <fct>  <dbl> <dbl> <dbl>
## 1 Female  29.9  67.3  14.5
## 2 Male    28.8  58.8  15.1
s3 = ggplot(df1, aes(x=BMXBMI, y=Systolic_BP1, color = Smoked_100)) + 
  geom_point(alpha=0.2) + 
  scale_color_manual(breaks = c('Yes', 'No'), values = c('blue', 'red'))
s3

corr_func('Systolic_BP1', 'BMXBMI')
## [1] 0.1352012
s4 = ggplot(df1, aes(x=Age, y=Systolic_BP1, color = Smoked_100)) + 
  geom_point(alpha=0.2) + 
  scale_color_manual(breaks = c('Yes', 'No'), values = c('blue', 'red'))
s4

corr_func('Age', 'Systolic_BP1')
## [1] 0.4692335
grid.arrange(
  s1, s2, s3, s4, nrow=2, ncol = 2, top = 'Testing some correlations'
)

Shiny dashboard with interactive scatter plot and box plot. Interactive table using the prop_func

Hypothesis testing - all below tests based on significance level of 0.05

T-test to see if male who has smoked 100 cigarettes in his live and had at least 12 alcoholic drinks in 1 year has a higher Systolic blood pressure than a male who has not. H0 != H1

test1_1 = filter(nhanes2, Gender=='Male' & Smoked_100=='Yes' & Alcohol_Year=='Yes')
test1_2 = filter(nhanes2, Gender=='Male' & Smoked_100=='No' & Alcohol_Year=='No')

#Test using the t.test function
t.test(test1_1$Systolic_BP1, test1_2$Systolic_BP1, alternative = "two.sided", 
       var.equal = FALSE, paired = FALSE)
## 
##  Welch Two Sample t-test
## 
## data:  test1_1$Systolic_BP1 and test1_2$Systolic_BP1
## t = 4.6317, df = 692.33, p-value = 4.333e-06
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  2.701071 6.676093
## sample estimates:
## mean of x mean of y 
##  128.8670  124.1784
#Test manually. As the std_dev on each sample are not similar, I have used the 
#unpooled approach
t1_n1 = nrow(test1_1)
t1_n2 = nrow(test1_2)
t1_u1 = mean(test1_1$Systolic_BP1, na.rm = TRUE)
t1_u2 = mean(test1_2$Systolic_BP1, na.rm = TRUE)
t1_sig1 = sd(test1_1$Systolic_BP1, na.rm = TRUE)
t1_sig2 = sd(test1_2$Systolic_BP1, na.rm = TRUE)

t_test = (t1_u1-t1_u2)/sqrt(((t1_sig1^2)/t1_n1)+((t1_sig2^2)/t1_n2))
t_test
## [1] 4.730432
#t-score for inbuilt_funtion = 4.6317, t-score worked out manually is 4.73. Both give a p-value of <0.00001 which provides us with sufficient evidence to reject the null hypothesis and say that a male who has drank and smoked does have a higher systolic blood pressure than a male who has not. We can futher back this up by looking at the confidence intervals (2.7, 6.67) and seeing that 0 does not fall within this. Will run similar test for only the alcohol part of this (below)

#T-test to see if somoen who has at least 12 alcoholic drinks in a year has a higher Systolic BP than someone who has not.

test2_1 = filter(nhanes2, Gender=='Male' & Alcohol_Year=='Yes')
test2_2 = filter(nhanes2, Gender=='Male' & Alcohol_Year=='No')

#As the variance is similar (testing below), we will use thr pooled approach
sd(test2_1$Systolic_BP1, na.rm = TRUE)
## [1] 17.61412
sd(test2_2$Systolic_BP1, na.rm = TRUE)
## [1] 17.8955
t.test(test2_1$Systolic_BP1, test2_2$Systolic_BP1, alternative = 'two.sided', var.equal = TRUE)
## 
##  Two Sample t-test
## 
## data:  test2_1$Systolic_BP1 and test2_2$Systolic_BP1
## t = 2.275, df = 2446, p-value = 0.02299
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  0.2770041 3.7363882
## sample estimates:
## mean of x mean of y 
##  127.4336  125.4269
#P-value of 0.02299 still provides us with enough evidence to reject the null and conclude that 'on average' people who drink 12 alcoholic drinks per year have a higher systolic BP than those tho don't

#Hypothesis test 3 - Do men of all ages have higher Systolic BP than women.
test3_1 = filter(nhanes2, Gender == 'Male')
test3_2 = filter(nhanes2, Gender == 'Female')

#Testing variance below. As the variance differs, we will use the unpooled approach
sd(test3_1$Systolic_BP1, na.rm = TRUE)
## [1] 17.64247
sd(test3_2$Systolic_BP1, na.rm = TRUE)
## [1] 19.06579
t.test(test3_1$Systolic_BP1, test3_2$Systolic_BP1, alternative = 'two.sided', var.equal = FALSE)
## 
##  Welch Two Sample t-test
## 
## data:  test3_1$Systolic_BP1 and test3_2$Systolic_BP1
## t = 7.4453, df = 5397.1, p-value = 1.12e-13
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  2.739755 4.698244
## sample estimates:
## mean of x mean of y 
##  126.9989  123.2799
#Based on this low p-value we reject the null hypothesis.

#Hypothesis test to see if women >69 have a higher Systolic BP than men of the similar age
test4_1 = filter(nhanes2, Gender == 'Female' & Age > 69)
test4_2 = filter(nhanes2, Gender == 'Male' & Age > 69)

#Testing variancwe below. As the variance is similar, the pooled approach will be used
sd(test4_1$Systolic_BP1, na.rm = TRUE)
## [1] 20.7647
sd(test4_2$Systolic_BP1, na.rm = TRUE)
## [1] 20.46368
t.test(test4_1$Systolic_BP1, test4_2$Systolic_BP1, alternative = 'two.sided', var.equal = TRUE)
## 
##  Two Sample t-test
## 
## data:  test4_1$Systolic_BP1 and test4_2$Systolic_BP1
## t = 2.4618, df = 851, p-value = 0.01402
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  0.7044388 6.2458476
## sample estimates:
## mean of x mean of y 
##  139.9763  136.5012
#Hypothesis test to see if females living alone earn more than males living alone.
test5_1 = filter(nhanes2, Gender == 'Female' & Household_Size == '1')
test5_2 = filter(nhanes2, Gender == 'Male' & Household_Size == '1')

#Signifincant difference in variance, so we will used the unpooled approach
sd(test5_1$Income_to_Pov, na.rm = TRUE)
## [1] 1.550615
sd(test5_2$Income_to_Pov, na.rm = TRUE)
## [1] 1.598473
t.test(test5_1$Income_to_Pov, test5_2$Income_to_Pov, alternative = 'two.sided', var.equal = FALSE)
## 
##  Welch Two Sample t-test
## 
## data:  test5_1$Income_to_Pov and test5_2$Income_to_Pov
## t = 0.11781, df = 685.79, p-value = 0.9063
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.2201615  0.2482691
## sample estimates:
## mean of x mean of y 
##  2.279837  2.265783
#With a p-value of 0.9063, we fail to reject the null hypothesis based on a significance level of 0.05. Confidence level also includes 0.

After some further analysis using the created Shiny dashboard, something interesting was found. Using the interactive box plots we can see that someone who has smoked 100 cigarettes in their life will on average earn less than someone who hasn’t. Even more interesting, someone who drinks at least 12 alcoholic drinks per year will earn more on average than somoene who has not (illistrated by the first 2 plots below). Finally the 3rd box plot shows that someone who drinks 12 alcoholic beverages in a year and who has never smoked 100 has a significantly higher income to poverty ratio (on average) than someone who smoked and does not drink. Based on these finding a hypothesis test will be carried out based on the first 2 plots to see if this is significant enough.

df1 = filter(nhanes2, Smoked_100 !="Don't know" & Smoked_100 !='Refused' & Alcohol_Year!="Don't know" & Smoked_100 !='Refused')
b5 = ggplot(df1, aes(x=Smoked_100, y=Income_to_Pov, color=Gender)) + 
  geom_boxplot()
b5

b6 = ggplot(df1, aes(x=Alcohol_Year, y=Income_to_Pov, color=Gender)) + 
  geom_boxplot()
b6

b7 = ggplot(df1, aes(x=Alcohol_Year, y=Income_to_Pov, color=Smoked_100)) + 
  geom_boxplot()
b7

2 hypothesis tests. First to see if someone who has smoked 100 earns less than someone who has not, and second to see if someone who drinks 12 in a year earns more than somone who does not. If the results fail to reject the null, a further test will be carried out to see is someone who has not smoked 100 but who does drink at least 12 alcoholic beverages in a year earns more than somoen who smoked 100 and does not drink.

test6_1 = filter(nhanes2, Smoked_100 == 'Yes')
test6_2 = filter(nhanes2, Smoked_100 == 'No')

#Difference in variance, so we will used the unpooled approach
sd(test6_1$Income_to_Pov, na.rm = TRUE)
## [1] 1.537442
sd(test6_2$Income_to_Pov, na.rm = TRUE)
## [1] 1.632999
t.test(test6_1$Income_to_Pov, test6_2$Income_to_Pov, alternative = 'two.sided', var.equal = FALSE)
## 
##  Welch Two Sample t-test
## 
## data:  test6_1$Income_to_Pov and test6_2$Income_to_Pov
## t = -6.8032, df = 4668.2, p-value = 1.152e-11
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.3926522 -0.2169758
## sample estimates:
## mean of x mean of y 
##  2.222975  2.527789
test7_1 = filter(nhanes2, Alcohol_Year == 'Yes')
test7_2 = filter(nhanes2, Alcohol_Year == 'No')

#Difference in variance, so we will used the unpooled approach
sd(test7_1$Income_to_Pov, na.rm = TRUE)
## [1] 1.619916
sd(test7_2$Income_to_Pov, na.rm = TRUE)
## [1] 1.502718
t.test(test7_1$Income_to_Pov, test7_2$Income_to_Pov, alternative = 'two.sided', var.equal = FALSE)
## 
##  Welch Two Sample t-test
## 
## data:  test7_1$Income_to_Pov and test7_2$Income_to_Pov
## t = 9.5149, df = 3179.7, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  0.3639530 0.5528841
## sample estimates:
## mean of x mean of y 
##  2.548941  2.090523

Both of the hypothesis tests reject the null, therefore no need for the third test. With a significance level of 0.05 we can say ‘on average’ someone who has smoked 100 cigarettes in their life earns less than someone who has not. We can also say that somone who drinks at least 12 alcoholic beverages a year will ‘on average’ earn more than somoen who does not.

Below box polts for further testing to see if Income, Education and/or Household_Size has on Systolic Blood Pressure

b8 = ggplot(nhanes2, aes(x=Income_Group, y=Systolic_BP1, color=Gender)) + 
  geom_boxplot()
b8

b9 = ggplot(nhanes2, aes(x=EducationX, y=Systolic_BP1, color=Gender)) + 
  geom_boxplot()
b9

b10 = ggplot(nhanes2, aes(x=Household_Size, y=Systolic_BP1, color=Gender)) + 
  geom_boxplot()
b10

Box plots indicated little effect income has on Systolic BP. However, both education and household size seem to have an effect.

Below 2 hypothesis tests. First to see if someone with a lower education has a higher systolic blood pressure than someone with a higher education. Second to see if someone living by themselves has a higher systolic blood pressure than someone with a household size of 2.

test8_1 = filter(nhanes2, EducationX == '< 9th grade')
test8_2 = filter(nhanes2, EducationX == 'College/Uni graduate or above')

#Difference in variance, so we will used the unpooled approach
sd(test8_1$Systolic_BP1, na.rm = TRUE)
## [1] 19.70467
sd(test8_2$Systolic_BP1, na.rm = TRUE)
## [1] 16.56066
t.test(test8_1$Income_to_Pov, test8_2$Income_to_Pov, alternative = 'two.sided', var.equal = FALSE)
## 
##  Welch Two Sample t-test
## 
## data:  test8_1$Income_to_Pov and test8_2$Income_to_Pov
## t = -36.99, df = 1564.9, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -2.385345 -2.145109
## sample estimates:
## mean of x mean of y 
##  1.281756  3.546983
test9_1 = filter(nhanes2, Household_Size == '1')
test9_2 = filter(nhanes2, Household_Size == '2')

#Variancxe very similar, so we will used the pooled approach
sd(test9_1$Systolic_BP1, na.rm = TRUE)
## [1] 19.89733
sd(test9_2$Systolic_BP1, na.rm = TRUE)
## [1] 19.31388
t.test(test9_1$Income_to_Pov, test9_2$Income_to_Pov, alternative = 'two.sided', var.equal = TRUE)
## 
##  Two Sample t-test
## 
## data:  test9_1$Income_to_Pov and test9_2$Income_to_Pov
## t = -6.9733, df = 2102, p-value = 4.132e-12
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.6781463 -0.3804397
## sample estimates:
## mean of x mean of y 
##  2.273171  2.802464

Both hypothesis tests allow us to reject the null and conclude that ‘on average’ someone with a lower education has a higher systolic BP than someone with a higher education and also that someone living alone has a higher systolic BP than someone with a household of 2